Finding the height of a ball with a geometric series

Homework Statement

A ball is dropped from one yard and come backs up $\dfrac{2}{3}$ of the way up and then back down. It comes back and $\dfrac{4}{9}$ of the way. It continues this such that the sum of the vertical distance traveled by the ball is is given by the series $1+2\cdot\dfrac{2}{3}+2\cdot\dfrac{4}{9}+2\cdot\dfrac{8}{27}+\cdot \cdot \cdot=1+2(\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{8}{27}+\cdot\cdot\cdot(\dfrac{2}{3})^n$). Find the height of the tenth rebound and the distance traveled by the ball after it touches the ground for the tenth time.

Homework Equations

$S_n=\dfrac{a(1-r^{n})}{1-r}$

The Attempt at a Solution

I know that the height of the tenth rebound is simply the tenth term in the sequence so $h=s_{10}=(\dfrac{2}{3})^{10}\approx0.0173$ yards. Now I thought that the vertical distance would be $1+2\cdot\dfrac{\frac{2}{3}(1-(\frac{2}{3})^{10})}{1-\frac{2}{3}}\approx1.96$ yards using the formula for a geometric series $\dfrac{a(1-r^{n})}{1-r}$. However, the book tells me that the answer should be $6\cdot(\dfrac{2}{3})^{10}\approx0.104$ yards. Now on the chance that I did misinterpret the book and the author meant the vertical distance the ball traveled the tenth time it hits the ground, shouldn't it be $2\cdot\dfrac{2}{3}^{10}$?

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HallsofIvy
Your series starts with $1= (2/3)^0$, with n= 0, not 1. The "10th" term is n= 9, not 10.
Your series starts with $1= (2/3)^0$, with n= 0, not 1. The "10th" term is n= 9, not 10.
I guess I should've stated that the textbook focuses on the series in the parenthesis because the answer for the height is indeed $(\dfrac{2}{3})^{10}$